Solve this

considering the given functions,

$x=a \sin 2 t(1+\cos 2 t)$ and $y=b \cos 2 t(1-\cos 2 t)$

rewriting the above equations,

$x=a \sin 2 t+\frac{a}{2} \sin 4 t$

differentiating bove function with respect to $t$, we have,

$\frac{d x}{d t}=2 a \cos 2 t+2 a \cos 4 t$            .....(1)

$y=b \cos 2 t-b \cos ^{2} 2 t$

differentiating above function with respect to t, we have,

$\frac{d y}{d t}=-2 b \sin 2 t+2 b \cos 2 t \sin 2 t==-2 b \sin 2 t+2 b \sin 4 t$      .....(2)

$\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d t}}=\frac{-2 b \sin 2 t+2 b \sin 4 t}{2 a \cos 2 t+2 a \cos 4 t} \mid$ from equation 1 and 2

At $t=\frac{\pi}{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{b}}{\mathrm{a}}$